text:
How can I get back up into flying after not flying for over 20 years?
I'm a former USAF pilot and former Bonanza owner. Haven't flown private in more than 20 years, but looking to get back into it. A friend who owns a 182 and will fly with me and I know some instructors. I don't even know where my old certificate and log book is, so this is really starting from scratch. What steps do you recommend?

Welcome to aviation.SE! I'm not sure there's a definitive answer to this, but AOPA's rusty pilot seminars are intended exactly for people in your situation. You might also like to take the [tour] if you aren't familiar with this site.

It's like riding a bike.  A friend of mine just jumped back in to aviation cold and only needed about 10hrs to complete his commercial single add-on.  Make a few flights in a 182, get a medical (or take the BasicMed route), and do a BFR with someone who is not a friend of yours (read: will give you honest feedback).  Set aside 10hrs for that BFR.  Some of the rules have changed but all of the basics are still pretty much what existed in 1997...except now paper charts are a waste of time.  Get the Avare charting app (free, no ads) if you use Android.

The FAA just published a safety briefing on this topic

I do not have a sanctioned answer, but I can give you my take on an overview of what I would plan for someone in your situation.
Your certificate duplicate can be ordered.
You will need a through review of airspace (things have changed), weather gathering (also changed), AIM stuff (changed runway markings, phraseology, etc.).  I would urge FAAST courses for various topics you might feel you can get brushed up in.
On the air side, if we were flying, we would do a "checkout" in the 182, covering the systems, cowl flap operation, targeted power settings, trim usage, etc.  A little airwork, steep turns, turning to headings, power on and off stall recoveries, emergency procedures and landings, takeoffs and go arounds.
Everyone is different, but the last guy, similar to you, flew C208 and stopped in 2003, and did not fly since then.  This summer he hit the books, watched a few videos, took a practice commercial written, and flew 3.4 hours with me.  I was real comfortable signing him off, and his wife, who was the concerned party, was real happy with the results.
Two months later we did a instrument proficiency ride, and that was a little more difficult for him, but he completed the ride in two flights and less than 4 hours.
Good luck and enjoy. Have your instructor put a new high performance endorsement in your new logbook for the 182.

Thanks, Mongo. I'm sure there are a few gremlins out there waiting for me. I plan on taking my time and listening to whomever is in charge.

Just doing a ride with a retired 757 driver who has not flown in 14 years.  He decided that he would get a Saratoga. There is slightly more complexity than in the 182, but it's not much. The biggest thing is actually accomplished studying at home.  There are simulators for much of the avionics, although the platforms for those simulators may be a pain.  (eg Garmin 430 sim runs on XP)  Have fun!

BTW, the ride with the Saratoga owner, who used to fly 757s took 3.5 dual, and 4 hours ground, and included a FR and an IPC.  The IPC took longer than the flight review. He did one flight in a glider since retiring 14 years ago.  So it can come back pretty quickly.

As long as you hold at least an FAA PPL, it should be fairly easy.  As a former military pilot you could easily obtain an FAA CPL in the category and class of aircraft you flew in the USAF per §61.73.
After that, it’s just a matter of regaining currency.  You will have to undergo a flight review with a CFI and receive a logbook endorsement per §61.56 plus an Instrument Proficiency Check, if you intend to fly under IFR, as required by §91.1069 and §61.157.  It may take a few flights with an instructor, but it will all come back to you sooner or later.
I would recommend reviewing your training literature you have from the USAF or obtain copies of the FAA’s Airplane Flying Handbook, The Pilot’s Handbook of Aeronautical Knowledge, The Instrument Flying Handbook, and The Instrument Procedures available to you for self study.  I would also recommend that you obtain the latest copy of the FAR/AIM for 2018 and keep that on hand for reference during these flight reviews.
If you don’t currently have access to your certificate, but know you possess one, contact the FAA and request an duplicate copy from them.  I’d guess the same kind of inquiry could be made with the USAF for your pilot qualifications with them.
Oh and one other thing:  did you obtain a military logbook or a civilian rating just prior to giving up flying?  You may see if you can obtain the paperwork from the FAA which you filed for these ratings.  These days everything in done on the FAA’s Integrated Airman Certification and Rating Application (IACRA) database, but any Aeronautical experience you file for application for a rating can be used as an official documentation of logged flight times.  I’d check on that to rebuild your logbook on.

You may also need to obtain a new Pilot License. A rule change went into effect in 2006, that by 2010 all pilots needed a plastic license, looks kind of like a credit card, with hologram on the front, mag stripe on the back, place to sign your name. 2010 was extended to 2013.  Can request it at FAA.gov. $2 for a duplicate, free if you want your license # changed from your SSN.

Apart from the flight training that other answers have covered very nicely, you may need a new medical certificate. You said you used to own a Bonanza so I'll assume you already hold an FAA private certificate (at least). Your options are:

Apply for a regular third-class medical. If you're in good health that should be straightforward, but if you have any health concerns then it would be worth talking to AOPA or an AME before you actually apply for the medical examination.
If you held any valid FAA medical between July 15, 2006 and July 15, 2016 then you could fly without a medical under BasicMed. AOPA has a lot of detailed information on it that's worth checking out (and see this question).
Fly a light-sport aircraft without any medical at all, using sport pilot privileges. (You don't need a medical for gliders or ultralights either.) You mentioned a 182 so I guess you don't plan to go this route, but it's worth including for completeness.


text:
What is the relation between maximum airspeed and engine power?
I know that the drag force on an airplane scales as (velocity)^2 which implies that doubling the airspeed will produce a 4x increase in the drag force. So my question is:
Imagine we have an airplane that flies 100MPH on 100HP. If we replace the engine with one that produces twice the power (200HP) I would anticipate that, everything else being equal, the most speed increase I could possibly expect would be (100MPH) x sqrt(2) or about 140MPH- and that if I wanted to double the airspeed I would need a 400HP engine. Is this correct?

In theory, yes. However, everything else isn't equal.
When you double the speed, you probably exceed the design parameters of the airframe. You'll definitely exceed the VNE of a 100mph airframe if you operate it at 200mph, and then all sorts of ugly things can start happening, from weird handling characteristics to large pieces, like the tail, coming off inflight. Remember that 4x drag means 4x the strain on the airframe. Is your airframe up to the task?
Most 100mph aircraft have a design optimized for short takeoff and landing, wings that are very efficient at slow speed, but not efficient at higher speeds. So the simple drag calculation may not hold true, when you take wings designed for slow speed and operate them at a much higher speed. The actual drag will probably be higher, if you consider the design of the airframe.
Compare a 100mph aircraft, say a Cessna 150, to a Mooney 201, that can go 200mph. The Mooney has much thinner wings, that result in a high landing speed, but lower drag at higher speeds, while the 150 has a fatter wing that gives a lower landing speed and better low speed handling, but may not fare so well at a much higher speed than it was designed to handle. The other tradeoff, aside from speed, is the 150 is far more forgiving of a clumsy pilot, while the Mooney is probably not the best trainer for first time students.
Note that the Cessna goes 100mph on 100hp, while the Mooney goes 200mph on 200hp, so the advantages of designing for the target speed can be seen in not just upping the power, but redesigning the airframe (and adding retractable gear).

OP stated: same aircraft.

The question was not about practical considerations, but about theoretical implications.

It isn't a valid theory if it can be easily disproven. You can't just consider power when doubling the speed of the aircraft. You have to beef up the airframe, or redesign it unless you want to risk catastrophic failure, which adds weight, which requires even more power. Plus, the wings will probably produce more drag than the calculations provide, because they were designed for lower speeds.

Your thinking is correct if thrust won’t change with speed. But it does.
Since you prescribe power, I have to assume a piston or turboprop engine. In that case, thrust is power divided by speed and your thrust will decrease with the inverse of speed. Not only will drag grow, but thrust will decrease at the same time, which will give you a speed increase of only 1.26 when you double engine power.
Mathematically speaking: $$P = \frac{T\cdot v}{\eta_{Prop}} = \frac{\rho\cdot S_{ref}\cdot c_D}{2\cdot\eta_{Prop}}\cdot v^3 \Rightarrow v = \sqrt[\LARGE{3}]{\frac{P\cdot\eta_{Prop}}{\frac{\rho}{2}\cdot S_{ref}\cdot c_D}}$$
I'm afraid that just swapping engines will result in a poorly matched propeller, so your $\eta_{Prop}$ will also decrease. Now much depends on the lift coefficient at which the 100 hp version flew. Flying faster at the same density will lower your lift coefficient and with it the induced drag, so the drag coefficient at the higher speed will be lower. By how much depends on the initial lift coefficient. If an underpowered aircraft gains a bit more power, the possible speed increase can indeed be proportional to the square root of the power increase. Normally, however, induced drag is almost negligible at the top speed of powerful propeller airplanes and the cube root law reflects reality best.
You will need approximately an 800 HP engine and matching propeller (ideally of the same mass as the 100 HP engine and propeller) to fly twice as fast.
Nomenclature:
$P\;\;\;\;\;\;\:$engine power
$T\;\;\;\;\;\;\:$thrust
$v\;\;\;\;\;\;\;$flight speed
$\rho\;\;\;\;\;\;\;$air density
$c_D\;\;\;\;\;$drag coefficient
$S_{ref}\;\;\;$reference area

But the thrust of a rocket, for example, stays the same at any speed, because the speed of the ejected gases is relative to the rocket itself, not to the 'stationary' observer... The same should be the case of the air accelerated by a propeller.

I've been watching a debate on the physics stack exchange about whether the horsepower requirement scales with the square, the cube, or the 4th power of airspeed. I thought it was with the square...

@xxavier Yes, but do you know a rocket where thrust is specified in horsepower?

@nielsnielsen … which only demonstrates that you better come here for the correct answer.

@Peter Kämpf  I'm not talking about horsepower, but just commenting on your statement that the thrust decreases with the inverse of speed. In my opinion, that statement is not correct...

@xxavier Well, you are entitled to your opinion, but I have read the question. It specifically asks for a power increase from 100 to 200 HP. Ergo, this cannot be a rocket, or a jet for that matter.

@qqjkztd: … at what speed? At what propeller speed? Agreed if flight speed is the same and propeller speed is 1.41 times higher without hitting transsonic speed at the tips. Then yes, thrust will double.

I went out to another website after googling airspeed versus horsepower and got the v^3 dependency quoted by Peter Kaempf, the derivation was a little easier to follow: drag scales as v squared, power is force x velocity, equate force with drag, there's your v cubed dependency. thanks to all, I enjoy reading your comments. -NN

@nielsnielsen: The question already gives the square dependency between drag and airspeed. Follow the links in the answer and every step is explained. I did not want to repeat everything again in this answer.

Let us continue this discussion in chat.

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text:
What is the low resistance design path in an aircraft to withstand a lightning strike?
I know in a building we have lightning rods that conduct a lightning strike into the earth, but for an aircarft there is no ground to 'ground' so I assume the lightning enters and passes through the aircraft. Therefore, since some areas must have high resistance paths of protection, for example a path through the antennas or main electronics bay, they must be an alternate low resistance path capable of conducting a very high power flow with minimum damage and burning. What is this design path?

The conductive exterior body
Almost all aircraft have metal bodies, the electric current flows through the exterior metal.
Aircraft with composite hulls can have conducting material added to the composites to provide a path for lightning. This is typically a metal mesh.
Some smaller composite aircraft, lacking this sort of provision, simply cannot fly near clouds or anywhere where there is a risk of lightning.

photo by Boeing - see references
The outer conducting surface acts as a Faraday cage preventing the high voltages and currents from affecting the interior.
 a Faraday cage photo by Antoine Taveneaux

It is estimated that on average, each airplane in the U.S. commercial
  fleet is struck lightly by lightning more than once each year.
Initially, the lightning will attach to an extremity such as the nose
  or wing tip. The airplane then flies through the lightning flash,
  which reattaches itself to the fuselage at other locations while the
  airplane is in the electric "circuit" between the cloud regions of
  opposite polarity. The current will travel through the conductive
  exterior skin and structures of the aircraft and exit off some other
  extremity, such as the tail. Pilots occasionally report temporary
  flickering of lights or short-lived interference with instruments.
Every circuit and piece of equipment that is critical or essential to
  the safe flight and landing of an aircraft must be verified by the
  manufacturers to be protected against lightning in accordance with
  regulations set by the Federal Aviation Administration (FAA) or a
  similar authority in the country of the aircraft's origin.

From Scientific American - What happens when lightning strikes an airplane?

Planes are now built to absorb 250,000 amps, whereas the average
  strike generates 32,000 amps.
"It only gets really serious when the radome [nose cone] is struck,
  the only part of the plane's shell not made of metal, as this is where
  the radar is located. But nose cones have special lightning conductors
  for just this reason."

From The Guardian - Should you be worried about lightning striking your plane?

The 787 flight test team gathered the unexpected data last month after
  one of the Dreamliner test aircaft was struck by lightning. Unlike
  traditional aluminum aircraft where the entire aircraft is conductive,
  on a composite airplane the charge from a lightning strike would find
  its way to the conductive parts such as wiring or hinges. In order to
  avoid the risk of the charge damaging these kinds of parts, Boeing
  had to add conductive material to the composites in order to provide a
  pathway for lightning strikes.

From Boeing 787 Withstands Lightning Strike

Other references

Boeing - Lightning Strikes, Protection, Inspection and Repair
Protecting the 787
Lightning strike protection strategies for composite aircraft

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text:
What is this hole above the intake of engines of Mil helicopters?
Almost all the helicopters designed and manufactured by Mil have a hole above the intake. e.g.

What is it for? APU? Other helicopters made by this company:

This link claims it's the central oil cooler intake for the Mi-26.

This is lubrication oil cooler.
APU intake is on right side and APU exhaust is on left side(small APU exhaust opening behind engine exhaust in first pic and third pic )

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Can I compute the lift coefficient based on the NACA airfoil?
I would like to know whether it is possible to compute the lift coefficient $C_l$ based on the NACA airfoil? Usually I look up the values on airfoiltools.com.
Eg: If I have a 2412NACA airfoil and would like to know the lift coefficient for an angle of attack $\alpha$ = 3°. What would be the corresponding lift coefficient?
Thanks

Be more specific, and an answer is possible: What is the Reynolds number and the Mach number? Do you need a 2D value or one for a specific wing?

Yes it is possible.
Here is what I did, maybe it answers your question.
1) I went to airfoiltools.com into the comparison section.
2) Looked up the NACA-airfoil 2412

3) scroll down to choose the speed (Reynoldsnumber). I took the minimum and maximum to get an idea of the spread.

4) scrolled further for the polars, at sketched the $\alpha = 3^\circ$ into it

But somehow I do wonder if this is really your question.
Or are you asking for xfoil and the commands to compute it?
In case you want to compute the values without visiting airfoiltools (which most-likely use xfoil themselves) you need a tools to numerically solve the governing equations.
You can do this in many ways. Those ways differ in the way the flow is approximated. So you could go with XFoil (very simple approximation of the aerodynamics) or maybe OPENFOAM (more complex description of aerodynamics).
There is however no way to calculate these coefficients in an easy way.

The thin airfoil theory approach (code in @Jacob Ivanov's answer) is detailed in this answer to a similar question.
However, in the link, the integrals have been carried out analytically -- vs. the provided code solves the integrals numerically.  The link works out the example for a 4412, but the equations can easily be re-evaluated for a 2412.

I was given this exact airfoil (NACA 2412) as part of an assignment for my graduate level Aerodynamics coursework, and since I already solved it, I figured I'd post it for the benefit of all future students who found this page.
My solution technique comes from pages 358 and surrounding of Anderson's "Fundamentals of Aerodynamics".
import numpy as np
import scipy.integrate as integrate

alpha = np.linspace(-10, 10, 100) # deg
NACA = '2412'
c = 1.2 # m

def z_prime(x, naca, c):
   m = float(naca[0]) / 100
   p = float(naca[1]) * c / 10

   if 0 <= x / c <= p:
      return (2 * m) * (p - x) / (p ** 2)
   if p < x / c <= 1:
      return (2 * m) * (p - x) / ((1 - p) ** 2)

def integrand_0(x, naca, c):
   return 2 * z_prime(x, naca, c)/ c / np.sqrt(1 - (2 * x / c - 1) ** 2)

def integrand_1(x, naca, c):
   return 2 * z_prime(x, naca, c) * (1 - 2 * x / c) / c / np.sqrt(1 - (2 * x / c - 1) ** 2)

def integrand_2(x, naca, c):
   return 4 * z_prime(x, naca, c) * (1 - 2 * x / c) / c / np.sqrt(1 - (2 * x / c - 1) ** 2)

A0 = np.zeros(len(alpha))
A1 = np.zeros(len(alpha))
A2 = np.zeros(len(alpha))

c_L = np.zeros(len(alpha))

for i in range(len(alpha)):
   A0[i] = np.deg2rad(alpha[i]) - integrate.quad(integrand_0, 0, c, args = (NACA, c))[0] / np.pi
   A1[i] = 2 * integrate.quad(integrand_1, 0, c, args = (NACA, c))[0] / np.pi
   A2[i] = 2 * integrate.quad(integrand_2, 0, c, args = (NACA, c))[0] / np.pi

   c_L[i] = np.pi * (2 * A0[i] + A1[i])

import matplotlib.pyplot as plt
plt.plot(alpha, c_L, label = "NACA {0}".format(NACA))
plt.axhline(0, linestyle = 'dashed')
plt.title("Coefficient of Lift over Angle of Attack")
plt.xlabel(r"$\alpha$ (deg)")
plt.ylabel(r"$c_L$")
plt.legend()
plt.show()


text:
What are the general problems faced for maintenance in fuel inerting systems?
Currently most of the nitrogen generation systems have Air Seperation Modules (ASM) and other associated mechanical components to generate Nitrogen Enriched Air.
Hence, are there any inherent problems with the system in terms of maintenance, weight or the controller?

I am asking about commercial airlines

As far as commercial airliners go, the inerting system is relatively simple and lightweight; see the below diagram from Airbus for a relatively representative architecture (most systems also have some kind of compressor to inert the tanks more quickly in certain situations).

With a total weight of just a few hundred pounds and the ability for most of the architecture to be colocated near the fuel tanks, the inerting system is far less complex and heavy than, say, the air-conditioning system. The air separation module (ASM) is practically the only unique component; most other maintenance issues with inerting systems are similar to those of other systems: mechanical/thermal fatigue and failure, software bugs, accidental damage, etc.
However, there have been some issues with the reliability of the ASMs, which is currently the biggest concern of aircraft operators. I can't find any public information on why so many ASMs are failing at half their predicted service life, but I imagine it has something to do with their being a relatively new mechanical technology on commercial aircraft.


text:
Do I need to get a new AFSP approval if I change my instructor or aircraft?
I plan to remain with the same provider though.

An AFSP application is valid for one provider, one aircraft category (TSA category, not FAA category), and one training event (single-engine, IR or multi-engine). If you change any of those things, you need to submit a new application.
If you're at a part 61 or 141 school that has multiple instructors, the approved provider is usually the school, not the individual instructor. So you can receive training from any instructor there without an issue.
If you're getting training directly from a freelance part 61 instructor, then it's possible that the instructor himself is the approved provider. In that case, you would need to submit a new application to get training from a different instructor or school.
In either case, your school or instructor should know exactly who's registered with the TSA as the provider.
As for the aircraft, you only need to submit a new request if you want to train in a different aircraft category. Initial training is usually done in category 3 aircraft (maximum MTOW of 12,500lbs) and you can train in one or many physical aircraft, as long as they're all in the same category. But remember that SEL and MEL are different training events so if your current approval is for SEL training you can't receive MEL training, even if the aircraft is under 12,500lbs.
Links to specific pages on the AFSP website don't work properly, but their FAQ is reasonably clear and you can email their helpdesk at AFSP.Help@dhs.gov if you need more information. AOPA has a useful guide, and the regulations (including category definitions) are in 49 CFR 1552.

So shifting between a Piper Warrior and a Tomahawk trainer should not be an issue, correct?

@Karthik Correct. They're both single-engine aircraft and less than 12,500lbs


text:
Would the front rudder of the Wright brothers really have helped in recovering the glider from a stall?
I have found a paragraph in the 1904 British patent of the Wright brothers (see the citation below) and it seems, from what they explain, that the two inventors believed the front rudder could have recovered the plane from a stall in any situation.
The question is, why were they so sure that after a stall the plane would have started to fall nose down, would have accelerated and the front rudder would have been able again to control the pitch of the flying machine?
If the main wings (and also the front rudder) had stalled their glider/airplane could have started to fall sideways or tail first.

"Contrary to the usual custom, we place the horizontal rudder in front
  of the main surfaces or “wings” at a negative angle, and use no
  horizontal tail at all. By this arrangement we obtain a forward
  surface which is almost free from pressure under ordinary conditions
  of flight, but which, even if not moved at all, becomes an efficient
  lifting surface whenever the speed of the machine is accidentally
  reduced very much below the normal, and thus largely counteracts that
  backward travel of the centre of pressure on the main surfaces or
  wings which has frequently been productive of serious injuries by
  causing the machine to turn downward and strike the ground head on."
   Wright brothers, “British Patent No. 6732, A.D. 1904 – Improvements in Aeronautical Machines”, Date of Application: 19th Mar., 1904, Accepted: 12th May, 1904.

  Wright glider

Even though they called it a "horizontal rudder" it's probably clearer if you call it an elevator or stabilizer.

Note the raised rear Vstab position (not only helps prevent  tail strike but also)  gives a little "top drag", which, along with more forward set CG, improves static stability.  This helps avoid stalling.  Now move forward apparatus to the rear (could be on top of the Vstab!) and ... now where do we put the high by-pass turbofans ... er ... um ... oh! ... propellers!

No, it would not have helped by itself.
What could prevent a stall or recover the aircraft from an incipient stall would be a counteracting movement of that canard surface, but that would need to be commanded by the pilot. It makes little difference if that control surface is ahead or behind the wing as long as it has enough lift potential left to produce the desired pitch down moment.
The way we understand stability today is that the aircraft will correct a deviation from the desired pitch attitude without pilot intervention. Using this definition, the Wright Flyers were unstable. Only by active and permanent corrections could straight flight be maintained. This is rather stressful and makes longer flights very tiring. Flying faster also can be ruled out because the speed of deviations goes up with flight speed, such that the human reaction time delays corrections too much and can even make them counterproductive.
The Wrights were correct that an unloaded surface gives them the highest control authority and, when moved in a correct and timely manner, helps to prevent a stall. However, placing that surface ahead of the wing comes with several disadvantages, loss of stability being only one of several. On the other hand, placing that lightly loaded surface at the rear results in a stable aircraft with better performance.

I will click to accept your explanation in about two weeks if no better answer is posted until then. Something else, I asked a different question (see: https://aviation.stackexchange.com/questions/73054/did-the-wright-brothers-discover-the-three-axis-control-system-for-airplanes-gli) and a user (Robert Werner) attempted an answer. Because you appear to have quite a lot of experience in aeronautics, I will dare to ask you whether his explanation seems valid or it is flawed. (Only if you have time to take a look at his answer, there is no emergency, no rush.)

@Simplex11: Robert knows much more about the Wrights than I, so I trust his explanations. The Wrights certainly were the first who successfully controlled an aircraft in all three axes; before them all elements had been developed already but not used in a practical application. Lilienthal already discussed wing warping, elevators and rudders with his fellow pioneers but his glider designs tied his hands (literally), so he stayed with weight shifting. Wolfmüller, one of those pioneers, put those ideas into a glider which is now on display in a museum. (https://tinyurl.com/yx4xtpzy).

They are describing the Flyer's canard configuration. The canard surface is the "horizontal rudder".  The wing airfoils they and others were using at the time had very strong pitching moments at stall AOA (due to the constant curvature or arc of the profile as you can see in the sketch), that would overpower the downforce of a horizontal tail, or even cause the tail itself to stall, and pitch the plane over violently.
The Wrights found that an up lifting surface at the front could deal with this (that is, resist the pitching moment of the main wing) better than a down lifting surface at the tail.  It was a solution to a problem they faced caused by the airfoil they were using.
The price of this was a very twitchy craft to fly in pitch because the canard had little in the way restorative forces itself due to the way it was mounted, so it had to be consciously held at a desired position, which they lived with and got used to.  Like with wing warping, they had little stability, but lots of control (too much in the pitch mode, you could say, as as result of making the canard big enough to resist the wing's pitching moment), which is what wowed observers, who were impressed with the control while unaware of the pilot's struggle with pitch stability when it was flying.
Later development of airfoils with better managed pitching moments (by moving most of the camber forward) made aft tails less problematic and the advantages of the aft down-lifting tail won out. The canard faded from the scene, only to reappear occasionally, with mixed success.

Several errors in your answer. 1:”the up force of a horizontal tail”; you probably meant horizontal stabilizer however the force is down ant not up. Is meant to pitch the plane up, which is not the same thing. 2:”down lifting surface at the front”; what is down lifting? You probably meant down pitching as pitching the nose down. 3:”down-lifting tail”, I have never heard of such a thing. Would you care to explain yourself?

"Up lifting force at the front... better than a down lifting surface at the tail".  A canard surface at the front lifts UP.  A regular tail at the back lifts DOWN.  What is the problem with that?  I don't understand what you are confused about.

The problem is that the canard pitches the nose down, otherwise the wing would pitch up and stall. This makes the basics of canard. The exact opposite of what you state. You can not say “lifts UP” because lifting mean getting something up and lifting up is redundant. You also say “a regular tail lifts DOWN”. Lifting something down is nonsensical because lifting means bringing up and down is the exact opposite. I hope you understand now.

A canard is an up lifting surface that supports perhaps 20% of the total weight of the machine. The horizontal tail on a conventional aircraft makes "lift" in the aerodynamic sense, not lift as in away from gravity; the "lift" is downward. A canard aircraft is like a 4 leg table with most of the weight on two legs at one end and a minority of weight at the other. A conventional aircraft is a see saw with the C of G on one side of the pivot and someone pushing DOWN on the other side of the pivot. He's not "lifting" in pushing down,but the force he is applying is "lift" in the aerodynamic sense.

We’re talking about the Wright brothers. The forward stabilizer had a negative angle and pitched the craft nose-down. As for canard, if you think dassault rafale  is supported 20% by nose stabilizer, then I don’t agree.

NO, the Wright British Patent No. 6732 appears to deal with post stall control of the aircraft.
Though 1904 A.D. is a long time ago, a modern wings center of lift still shifts backwards to the center of the wing after it stalls.  Even their curiously proto laminar/Davis design thin undercambered wings had a center of lift further forward at lower AOA "unstalled".
Notice you don't see an airspeed indicator on that aircraft.  At this early stage, stalling was a frequent occurrence, and, at low altitudes, would result in a ground impact.
The design only had value in perhaps softening the blow from a nose in to a belly flop post stall impact, but actually hindered stall recovery.
As designers began to better understand proper relationships of center of pressure to center of gravity, then they came to realize that this post stall pitching down behavior actually helped unstall the wing faster, helped even more by placing the horizontal stabilizer in the back of the aircraft, as the Wrights did on their successful Model B.
This lead to the modern solution to deal with stalling:  avoid it at low altitudes with airspeed and AOA instruments, and pitch down to unstall your aircraft, which works at altitude.  Simply do not accept even controlled crashing as part of your flight envelope.

What they had was a further called canard configuration with the horizontal stabilizer ahead of the wing. They stated that the stabilizer is angled negatively and not loaded in normal conditions.
The center or pressure is located 1:3 of the wing cord towards the leading edge. For this reason in flight the wing has a tendency to tilt up, pitching the nose of the craft towards the sky. If left uncompensated by the horizontal stabilizer, the wing will fly at high angle of attack, will loose speed and stall.
To avert that situation, they devised the forward stabilizer meant to forcibly keep the nose down. Situation in which the wing gains speed, lift increase, the wing pitches (the nose) up, the drag increases, the wing slows down, the lift decreases, the nose pitches down. They must have optimized the design to the point where these nose oscillations were canceled at least in normal flight.
About horizontal stabilizers
The positive angle of the aft stabilizer contributes to maintaining the nose down after the wing stalled. This must have been known by Wright brothers.
The negative angle of the forward stabilizer used by Wright brothers maintains the nose down before the wing stalled, practically inviting the wing to speed up. However the wing would pitch up while speeding, which in the event of overcoming the forward stabilizer, will lead to higher drag and slow down.
As long as the wing was not able to change speeds too fast, the self control of the pitch worked way better with the forward stabilizer.

What you say is correct only if the speed of the plane/glider is above what the Wrights called "normal". "Whenever the speed of the machine is accidentally reduced very much below the normal [I quoted them]", which means stall, the front elevator will also stall and will no longer be able to control the pitch of the flying machine. The plane can stall while following an ascendent course, while climbing. The nose is evidently up and the front elevator in stall. How can the glider be recovered?

The glider can’t recover from any stall. The reason is that there are stalls outside the flight envelope. The stall they were referring to is in level flight where the speed can’t sustain lift. If the speed is above normal but still in the envelope, there is no stall. Since the forward stabilizer is angled negatively it won’t allow high pitch, however if high pitch stall is met, I don’t think Wright brothers had a solution for it.


text:
What is the purpose of a fuse pin in a turbine engine?
what's the difference betweeen fuse pin and mounting bolts of an engine ? Are they the same ?
Any digram of a GE-90 engine showing those would be appreciated.

Fuse pins are basically metal cylinders which are designed to shear at a certain load. There is a fairly detailed description of the pins used on the Boeing 747 in this FAA report. Other Boeing airliners use similar pins.

Fuse pins were designed to let the pylons (and their engines) detach in case of wheel up landing, to prevent breaking the wings and igniting the fuel from the wing tanks. I'm not sure they are used today. There are shear pins at the engine mount level (the connection between the engine and the pylon), but not sure they play this role, it looks like they bear the lateral loads (thrust) in place of bolts. A shear pin is a pin, not a bolt. It cannot fix something onto something else, it does work in tension, only in shear.


text:
Dimensional analysis with wingspan length?
I need some help in understanding the fundamentals of dimensional analysis.
The following is is from "Fundamentals of aerodynamics" by John D Anderson, fifth edition:

Why is the wingspan length not added to the variables that the aerodynamics force R depend on?

you should avoid posting an image full of text. you should really copy that text here, at least

the text has lots of special texts and formulars, by the way I want to keep it original

yes, right. the book is quite popular and it's fifth editon

@DatXaLin You can use MathJax to typeset the formulas. The image is inaccessible to search engines and the partially sighted and neither of those things is good for the site.

The body is of a give shape and of a give angle of attack. The only variable left is then the scale of the body.
The author decided to use the chord length as reference length; if the shape is known and the chord length is known, all dimensions of the aircraft are follow from that.  Alternatively he could have chosen the wingspan, tail height or any other length as reference length.

I don't think so. Given shape only provides us the sectional airfoil and the chord length only give us the scale of this sectional, not how long the wingspan is.

If the body is of a given shape, then the wingspan will X times the chord length. X will be dimensionless.

thanks, if so, why X is not added to the variables. i am sure the aerodynamics force depend on X as well as c, V..

It my also depend on 2, 0.5, $\pi$ and other dimensionless numbers that are not included either. It is dimensional analysis after all.

Also X is not a variable, it is a constant for the given shape.

thank u so much, I think the author idea is that the wing need one parameter to determine it, like you said at first

Yes, exactly, that is the idea. You got it now :-)

In page 38 of the book you're quoting, the derivation goes on replacing $c^2$ with $S$,   the planform area of the wing, implicitly introducing the wingspan...

I have read to that page but it still does not help


text:
How is the mass flow between 2 streamlines derived?
I am reading "Fundamental of Aerodynamics " by J.D.Anderson Fifth edition, At pages 177 and 178 (section 2.14 Stream funciton), the book said: given 2D steady flow, from:
$$\frac{dy}{dx}=\frac{v}{u}$$
with u and v are known function of x and y, then we obtain f(x,y) = c by intergration.
If we have steamline 1 is f(x,y) = c1 and streamline 2 is f(x,y) = c2, then why is the mass flow between two streamlines equal (c2 - c1)? The book said that but there is no derivation of that.
As far as I can do:

Short answer: Because that's how we've defined $c_1$ and $c_2$.
Long answer: Read 2.14 carefully. As you've said, we have $f(x,y)=c$ via integration. Anderson then chooses $\bar{\psi}$ as $f$ (a convention) to get $\bar{\psi}=c_1$ and $\bar{\psi}=c_2$ for the two streamlines. However, we know that $c_1$ and $c_2$ are just arbitrary constants of integration that we can set to whatever we want. Anderson just happens to know that later down the road it will be helpful to

define the numerical value of $\bar{\psi}$ such that the difference $\Delta\bar{\psi}$ [between the two streamlines] is equal to the mass flow between the two streamlines.

Furthermore,

$\Delta\bar{\psi}$ equaling mass flow (per unit depth) between streamlines is natural. For a steady flow, the mass flow inside a given streamtube is constant along the tube; the mass flow across any cross section of the tube is the same. Since $\Delta\bar{\psi}$ is equal to this mass flow, then $\Delta\bar{\psi}$ itself is constant for a given streamtube.

OMG, I just found out that it's just defined ;)) I also just know that: with that definition, we can obtain the velocity field from such stream fuctions. Thanks


text:
Did the Westland Lysander have a helmet/face mask for radio communication during WW2?
I have a character in some war-time fiction I'm writing 'acquiring' an aircraft (a Westland Lysander).
Just need to know: was there a helmet/face mask for radio communication for these aircraft? And, by default, was it present in a 'parked' plane?
fwiw...my character's genned up with the entry, startup and takeoff procedure.

Since they were not suited for combat, they transitioned to supporting clandestine operations where, to quote the  Royal Aviation Museum of Western Canada,
To give it the long range it needed, the aircraft had to be lightened by removing all unnecessary equipment such as guns, armour protection and excess radio equipment, retaining only the radio/telephone for communication with the ground.
It appears that they did have radios, not sure about how they were used. But if you look at the video in the linked page at about 2:33 the pilot appears to be wearing a helmet with built-in headset and a face-mask type microphone. At 6:19 the rear crewmember looks like he is wearing a mask as well.
This is the same aircraft that features prominently in Code Name Verity by Elizabeth E. Wein.

re. the leather helmet/mask. I've found a still to corroborate this:
https://goo.gl/YWBiqc

It could also just be an intercom setup.


text:
Why isn´t TOGA thrust always 100% N1 or N2?
It makes sense to me, that under normal circumstances the engines shouldn´t be pushed to their maximum, to reduce wear and tear. But why isn´t TOGA thrust, which is the maximum selectable thrust setting, not corresponding to the maximum possible thrust, thereby 100% N1 or N2 whichever is the limiting factor.(In the picture TOGA is limited to 88.9% N1)

Does TOGA always correspond to maximum possible thrust? It makes sense for a go-around to use all the thrust you can get, because the aircraft is in a high drag, low lift configuration (flaps, slats, spoilers, landing gear, whatever else you've got to slow it down and force it toward the ground...) coming in for a landing and needing to regain altitude and speed as quickly as possible, but for take-off (other than the landing gear) the aircraft is in a low drag, high lift configuration already. Quite different scenarios...

Imagine you are on your takoff run, and suddenly there is an obstruction down the runway, you are above V1 so you can´t reject the takoff so what you wanna do is get maximum thrust to make it over the obstacle. So what you, as the pilot would do, is set the thrust from FLX to TOGA to hopfully make it. So even if you are on takeoff there are cases where you want the maximum thrustsetting to really be the maximum the engine can give you

Depending on the situation max power is limited by Vmcg/Vmca. If an engine fails at 100% the rudder may not have enough authority to counter the asymmetric thrust.

@TomMcW, that appears to be the correct answer here. The accepted one almost certainly isn't.

TOGA thrust is not always 100% N1 (or N2), because TOGA thrust is actually not about delivering the absolute maximum thrust that can be provided. Rather, it's about delivering a guaranteed large amount of thrust, to ensure the performance of the aircraft is what the pilot expects, and what the flight manual states will occur. The level of thrust needs to be predictable. So it's got to be the same with a new aircraft, as one 5 years old (for example), and the engines are about to be overhauled.
To achieve this, engines are "flat rated". See What is a Flat Rated Engine?

To repeat that answer a bit, a flat rated engine, produces a constant, or the same amount of thrust, up to a certain ambient temperature. This is achieved by having a margin (gap) between the maximum thrust the engine provides, and what it can really produce. This also results in a temperature margin between the engines actual turbine temperature, and the red line temperature. This means as the engine degrades, it can increase its operating temperature without hitting the red line (until it has degraded a lot), and so still provide the same level of thrust.
When a gas turbine's engine control is designed to produce the same thrust at different ambient temperatures, it means N1 speed will be higher on a hot day, than a cold day. See how in the figure, for a constant thrust, N1 is increasing as ambient temperature goes up? N1 has to go up as ambient temperature increases, because hotter air at the intake is less dense than colder air, so the engine needs to run at a faster RPM to make up for the lower density.

Flat rating is not to provide uniform performance—the performance calculation needs to be done before every take-off anyway—but to avoid exceeding other limiting factors inside the engine (since temperature is lower in that range, the limiting factor is likely pressure).

Flat rating is a responsibility of the FADEC. However, the pictured limit looks like a FMC one. That would most likely be due to Vmcg limitation that @TomMcW mentioned i nthe comment to the question.

@JanHudec, flat rating is to provide a TIT margin for deterioration. That causes engine performance to be independant of age, i.e. predictable. The pre take-off performance calc assumes this level of thrust - the calculation  is not tailored to each aircraft (by registration number).  If thrust needs to be reduced to avoid exceeding rudder authority, that is another (seperate, additional) factor. (As TomMcW says "depending on the situation")

The N1 and N2 values are simply RPM speeds, expressed as percentage of some nominal value. (It’s easier to talk about 95.6% N1 than 10,250 N1 RPMs.) But thrust for modern jet engines, and limitations for that thrust, are defined by the engineers as actual thrust, which is not a value directly observed on the flight deck. And due to factors such as altitude and temperature, you don’t have a set relationship between N1 and thrust. The thrust you get at 100% N1 in mile-high Denver on a hot summer day, you may get at 90% N1 at a sea-level airport mid-winter.
So a performance computer does the math & tells you what the N1 value is, given the conditions here & now, that will give you the specified thrust.
There is nothing special, or necessarily limiting (in many engines, at least) about “100%” N1. It isn’t uncommon on the CFM-56 to see N1 values above 100% for max takeoff thrust. That 100% is simply an RPM setting that matches the chosen nominal value, so if TOGA thrust today is 100.2% N1, that’s what you’ll set. In the photo in the OP, there is a redline (actually red arc) on the N1 indicator, but it is well above 100%.
N2, being engine core speed, has even less relationship to thrust than N1 does.

This doesn´t explain why the TOGA thrust setting isn´t always the maximum N1 RPM. There might not be a set relationship between N1 and thrust, but the both are definitely proportional.

TOGA power could be defined in various ways... by RPM or EGT or EPR, OR by the actual thrust being produced. Modern airliners performance uses the latter definition, and the pilots set the N1 that will in these conditions produce that XX,XXX lbs of thrust. TOGA is not “the most power the engine can possibly produce” (i.e. maximum N1 or EGT or EPR), but rather a set value. Performance charts (and engine lifespan) work better with a set value, rather than always having the engine going to “giv’in ‘er all she’s got, Capt’n” power (see Scotty in Star Trek).

Many pilots confuse 100% N1 as max thrust or 100% thrust.Although N1 always corresponds to a certain valve of thrust, In gas turbines 100% N1 means that the turbine is operating at or close to 100% efficiency.
5

I'm quite curious, do you have further reading about relationship between N1 and efficiency?

The assertion that 100% N1 = 100% efficiency is incorrect.


text:
How can a helicopter show up at 29000ft, 547kts at Flightradar24?
Today I observed a strange flight on Flightradar24, a helicopter departing HAM and traveling at altitudes up to 29000ft and speeds up to 547kts which seems to be way beyond the specifications of a helicopter.1
The altitude and speed were quite constant on this level between Hamburg (ATD 16:15) and Frankfurt (about 500km from Hamburg, after 30 minutes flight) and then it began descending and slowing down.

The details in normal resolution for better readability:

About 17:00 the symbol didn't move anymore and the helicopter was now down on 2875ft and 177kts near FKB (Karlsruhe Baden-Baden) and then disappeared.
I checked the registration and is indeed the Agosta Westland AW139 shown on the picture.
Does anyone have an idea how this could be. Usually, the data on Flightradar seems to be reliable, but a helicopter at 547kts seems odd.

1. According to Wikipedia the Agosta Westland AW139 can travel up to 306km/h (165kts) and has a service ceiling of 20000ft.
edit: speed was 547kts, not 247, added image with details.

FlightRadar data can have erroneous readings, I would not assume that it is correct. I would chalk this one up to bad readings.

In addition to what mins said, in that area and altitude today @17:00 there were ~90-100 kts of tailwind.

Obviously, the data is wrong, helicopters can't do this, you have explained it nicely yourself. What's the actual question?

thanks for the feedback. No doubt there's something wrong. I just try to understand how the system works and how the data can be wrong. Good hint with the Falcon 2000, that makes more sense but still wonder how the data got mixed up.

Wonder if it might be an issue with the Mode S registration -- somebody's database has a typo in it, perhaps?  Can you pull up other flights in FlightRadar24 of that same aircraft and see if they also look more Falcon-ish, or if everything else looks more like a helo?

FR24 is great but every so often there are bugs like this. I’ve heard of people making complaints based on what they’ve seen on the website, even reporting a missing aircraft (which had simply dropped out of their coverage). Rest assured that the real ATC has much better data than what’s on FR24!

Thanks Ben and Ralph. The recent data looks all more "falconish" than heli-like. For me it was just a surprise. I'm using FR24 occasionally  but didn't notice such mistakes. Now I take the data from FR24 with a grain of salt and read more about FR24 to better understand how it works.@all Thanks again  :)

Unrelated to the correct answer below, in the US, Some military flights have the ability to change the transponder code or incorrect flight information is provided to controllers regarding a flights capabilities. Military, police and ambulance helo's often only transmit mode C

@mins answer contains good information, however the conclusion that the mistake is in the Flightradar 24 database is not correct.
It is not Flightradar 24 that has a database error:

30040D IS I-PTFT it is NOT I-MOFI F2TH I-MOFI is miscoded on one box
  and using 30040D incorrectly. I-MOFI is 30047D please change it back!
  Why was it changed in the DB when I-MOFI was correct in the DB, I
  thought miscodes were not allowed anymore?

What this says, is that the 24 bit Aircraft Address (a.k.a. the Mode S address) 30040D belongs to the aircraft with the registration I-PTFT.
The registration I-MOFI has the 24 bit address 30047D, but the quote says it is miscoded on one box (transponder).
Flightradar 24 receives the Aircraft Address with the ADS-B messages from the aircraft and uses the database to find the aircraft associated with the address.
From the screenshot it can be seen that the address they receive is 30040D. In their database, this is correctly linked to the I-PTFT helicopter.

The fault is with the aircraft with registration I-MOFI; it transmits the 24-bit address 30040D while it should have transmitted 30047D. The transponder has the wrong 24-bit address encoded into it.

Thanks for that anwer! Turns out to be a more interesting case than I thought yesterday. So the transponder just uses this adress as an id and the registration and a/c type and other data are feteched from a database, right?!. Isn't that a serious thing that should be fixed asap? Isnt't it important for ATC?

It should be fixed, it is important to ATC, but is is not as serious as you might think. The 24 bit Aircraft Address is not used for aircraft / flight identification by Air Traffic Controllers.

Based on this discussion at FR24, mode S transponder addresses 0x30047D and 0x30040D were not associated with the correct airframes (as @Ralph early suggested).

30040D IS I-PTFT it is NOT I-MOFI F2TH I-MOFI is miscoded on one box
  and using 30040D incorrectly. I-MOFI is 30047D please change it back!
  Why was it changed in the DB when I-MOFI was correct in the DB, I
  thought miscodes were not allowed anymore?

Due to this address mistake when the aircraft is fetched by the tracking site, the wrong aircraft appears (I-PTFT instead of I-MOFI).
The actual aircraft is I-MOFI, a Dassault Falcon 2000:

I-MOFI at Salzburg Airport W. A. Mozart, photo by Karl Dittlbacher


text:
Can military personnel land their own private aircraft at military bases?
First off, I am asking similar, but logistically different question to What process do I follow to land a civilian aircraft on a military base?
If I am in the US Military (with a valid ID), am I free to land my personal (civilian style) airplane (with my civilian licence) at military bases - in the same sense that I could drive my car on?

I imagine the answer would be identical to the other question, except that instead of going directly to the base commander you'd submit your request through the chain of command, and special arrangements may be made so you don't have to notify the base commander 24 hours in advance of every intended landing...

Obviously that's an option.  But I wonder if you even need to do that as a member of the military, or if you can just show up.

32 CFR 855, Table 1 (linked in the answer) seems to indicate Service members, operating their own aircraft, leased aircraft, or other available aircraft for official duty travel (temporary duty, permanent change of station, etc.) or for private, non revenue flights are eligible for landing rights. It looks like they still need to file the paperwork, and with a few exceptions the request can be approved locally (at the base level).

You know, @voretaq7, if you want to put that into an answer.  It's definitely worthy.

I always tell passengers, when they ask if we can land at KNKX as we fly over it, that we can land anywhere you want to... once!

I seriously doubt you can drive your car onto any military installation you want without prior clearance. Same with an aircraft.

@jwenting - if you are military (what the question was about), you can certainly drive onto almost any military base with no prior clearance. The only exceptions would be the few high-security ones, like Groom Lake, etc.

^^^what he/she said.

@jwenting they USED TO have stickers on cars that allowed people to drive right on and get waved (or saluted) on without any hassle.  Now you need to show an ID, but that's about it.

@LarianLeQuella interesting, thought they'd have tighter security in place, especially after things like the Ford Hood shooting...

@jwenting in THAT particular case, it wouldn't have helped since the shooter was a person that was assigned to that base.

As someone already said, 32 CFR 855, Table 1 states:

Service members, operating their own aircraft, leased aircraft, or other available aircraft for official duty travel (temporary duty, permanent change of station, etc.) or for private, non revenue flights

This of course means you still need to contact base ops for a Prior Permission Required (PPR) of some sort (if required).  If the base has an Aero club, then your logisitcal considerations are much less (most aircraft probably don't want to refuel on JP-8).  The list of Aero Clubs seems to have decreased since I was in though.
The Flight Infomration Handbook will have all the base requirements listed in it.

Even military aircraft frequently need to get PPR's to land at other military fields.  It just ensures there is enough space and ground crew to handle the arriving aircraft.

Yes, military personnel can apply for a Civil Aircraft Landing Permit for official business.


text:
How much "power" do ATCs really have?
After reading this question about being cleared to go faster than 250kts in airspace that otherwise has a speed limit, I began to wonder how far can Air Traffic Controllers (ATCs) can bend the rules on an aircraft-to-aircraft basis?
Is there a limit to what exceptions ATC can make for someone?  I'm particularly interested for the answer in the USA, but I'm sure information about other countries would also be enlightening.

If you are using power in terms of authority, this question discusses it thoroughly and in detail.

On the other hand, if you're using power to mean power, it will vary a lot by the equipment under their control at a given field, but will usually probably be measured in kilowatts. :) Pilots, on the other hand, may have megawatts (GE90-115b comes in around 75 MW, IIRC.)

Related: http://aviation.stackexchange.com/questions/30134/can-a-pilot-take-off-at-his-own-discretion

As an active area radar controller, I completely agree with @rbp answer.
To better explain relationship between power and responsibility, I will try to describe what a controller would usually do: If I need an aircraft, for example, to reduce speed...
I know approx. what would be the minimum convenient speed for a particular type to ask, so I'd take this into account prior asking. But if a pilot tells me that he's unable to comply with the request, I'll find another solution i.e. radar vectoring to loose few miles or in extreme situation, level change. In short, a controller will never force a pilot to action if pilot answers that he can't do that. A controller must plan in advance that pilot may not be able to fulfil his request and have alternative solution. If you, as a controller, put yourself in a situation where you have no alternative, that is usually too late.
There are usually only few things where I can't make exception:

Minimum separation between aircraft
Danger/Prohibited zone
Terrain

Even then, in extreme case, I've never met: Imagine a situation where you have an aircraft which can't change level due to traffic above and below and has a crossing traffic. Then only remaining action I have available is radar vectoring, but what if the pilot advises me that his unable to turn due to weather, (the described situation is usually the controller fault because he should have predicted that earlier). Then controller probably would give the pilot traffic information (tell the pilot what to expect) and advise the minimum distance but it will not force the aircraft to CB.

Hello Sinisa Drpa, welcome to Aviation.SE!

As far as I know, most regs have some sort of out that says "unless directed by ATC," and ATC has two unbreakable rules which limit much of what they are allowed to do:

thou shall not vector airplanes into obstacles, terrain, unflyable weather or atmospheric conditions, restricted (in the lay sense of the term) airspace, or other aircraft
thou shall not deviate from aircraft separation

Other than that, when it comes to clearances, ATC has a fair amount of latitude to give you what they want, or what you ask for.
The balance of responsbility between ATC and the pilot in command, is covered by 14 CFR 91.123:

§ 91.123 Compliance with ATC clearances and instructions.
(a) When an ATC clearance has been obtained, no pilot in command may
  deviate from that clearance unless an amended clearance is obtained,
  an emergency exists, or the deviation is in response to a traffic
  alert and collision avoidance system resolution advisory. However,
  except in Class A airspace, a pilot may cancel an IFR flight plan if
  the operation is being conducted in VFR weather conditions. When a
  pilot is uncertain of an ATC clearance, that pilot shall immediately
  request clarification from ATC.
(b) Except in an emergency, no person may operate an aircraft contrary
  to an ATC instruction in an area in which air traffic control is
  exercised.
(c) Each pilot in command who, in an emergency, or in response to a
  traffic alert and collision avoidance system resolution advisory,
  deviates from an ATC clearance or instruction shall notify ATC of that
  deviation as soon as possible.
(d) Each pilot in command who (though not deviating from a rule of
  this subpart) is given priority by ATC in an emergency, shall submit a
  detailed report of that emergency within 48 hours to the manager of
  that ATC facility, if requested by ATC.
(e) Unless otherwise authorized by ATC, no person operating an
  aircraft may operate that aircraft according to any clearance or
  instruction that has been issued to the pilot of another aircraft for
  radar air traffic control purposes.

From what I've learned reading here (I am not a pilot), basically any instruction ATC gives can be overridden by the pilot, though, who has the ultimate responsibility for the safety of his plane. Is that correct?

@FreeMan i've updated my answer with the answer to your question

The ATC guys can't just do whatever they want. They have to operate within guidelines - they will have an ATC handbook/manual etc, much like a pilot has a flight operations manual, which has been merticulously developed and approved by the authorities. There are references in this manual as to when ATC can cancel speed restrictions, or deviate from the other usual procedures. In most cases it will read something like "Speed restrictions under 10,000 feet may be cancelled in Class C airspace only, and if the controller is satisfied that no risk of LOSA (loss of separation assurance) exists".
So in practical terms, ATC do have the ability to deviate from most of the usual procedures, but always within existing guidelines. Of course, everything goes out of the window in an emergency, however.

I concur with casey's answer on the linked question. There are two basic facts:

The FARs say what a pilot can or cannot do.
The FARs say a pilot must comply with an ATC instruction.

And then a complication:

ATC might instruct a pilot to do something specifically disallowed by the FARs (for example, exceed 250 KIAS below 10000 MSL).

Really what it comes down to, on both sides, is do what your career can handle. ATC shouldn't predicate separation on pilots agreeing to break a rule in the FARs, and pilots shouldn't break a rule just because ATC said so (unless they're certain that the rule is one of the "unless ATC says you can" rules). But if nothing bad happens the likelihood of anyone getting in trouble for it is minimal. Conversely if something bad does happen "ATC said I could" or "I didn't know pilots couldn't do that" might not carry a lot of weight in court. It's a judgement call.
I have seen anecdotal radio exchanges with pilots reporting obviously fabricated tailwinds when called out on the 250KT rule, or ATC snarkily commenting that two planes at the same altitude must have two different tailwinds.
The real bottom line is: the pilot-in-command is the absolute final authority as to the safe and legal operation of the aircraft, and "unable" is a perfectly valid response to any instruction so long as you can back it up with a good justification.

The speed limit of 250kts below 10,000 feet is mostly there to protect smaller aircraft and rotocraft.
For example, you might have a parachute jump plane flying around, and you don't want a Citation business jet bearing down on them (or the jumpers) at 400 knots: At that speed you don't have much time to react: You look away for two seconds to drink your coffee and BAM!
There was a famous accident a couple of years ago in which a glider pilot was flying around Reno and a private jet flew right into him (NTSB file LAX06FA277B) . That is exactly the kind of accident speed limits are supposed to mitigate.
If a pilot exceeds a speed limit, it really doesn't matter what ATC said, it is the pilot who is responsible. If the FAA determines he did something reckless it makes no difference that some controller told him it was ok.

Thanks.  But this answer is probably better suited for the question linked in my question, not to my question itself.

@SpongeBob What exactly is your question? If it is "authority" of an ATC to delete a below-10k speed limit in the US, personally I would say "none" because like I said, if the pilot exceeds the limit and crashes into somebody and the FAA determines he used poor judgement in going over the speed limit, what ATC said will make absolutely no difference.

I'm asking about their authority to alter ANY rule in the FAR, not just the speed limit.

@SpongeBob That is pretty broad. The FAR is like 1000 pages long. Just as a general principle my essential point would probably be valid, which is that an ATC instruction cannot immunize a pilot from a FAR violation.

The actual incident which lead to the 250-knot speed limit was a crash between a DC-8 and a Super Constellation. The CAB report on the incident is available from the FAA's website


text:
Why is 0 ft a valid target altitude for an autopilot?
Inspired by this story;

The pilot of a Flybe plane that dropped 500 ft (152m) in 18 seconds had
  selected the wrong autopilot setting shortly after take-off, an
  inquiry has found...
...The plane climbed to 1,500 ft, but then pitched and "descended
  rapidly" because the autopilot was set with a target altitude of 0 ft.

Why is 0 ft a valid target value for the autopilot? Is there any practical use for being able to set it so low and are there safety features that would prevent the plane just hitting the deck?

To cruise at low altitude in Death Valley?

@quietflyer OK, good point that the ground level might be below 0ft sea level, I hadn’t thought of that. But in my head I imagined there would be some kind of buffer below where the autopilot couldn’t be trusted.

Yes. me too, that's why I didn't form it into a real answer.  Good question.

Which plane was that? Dash8? Bae146 or ATR? After it will depend of the autopilot coupled with the FD. For Airbus and Boeing the minimum is 100ft. But the autopilot is capable to land the aircraft so, it is able to go below 0ft ASL if you land in Amsterdam for example.

@ChrisLau a Dash-8 Q400.

@757toga yes you are right. On Boeing and Airbus, you will have a compilation of egpws, rad alt, and ILS Location and GS. So it is targeting a trajectory and will do the flare when 50ft agl is reached.

Now for that story. If you are flying at LFBO when the altitude of the threshold on 32L is 512ft and your FCU is targeting 500ft. Your aircraft will go to the ground. The non precision of asl and agl makes that statement on targeting 0ft not consistent.

@757toga agreed.

The full AAIB report is here: https://assets.publishing.service.gov.uk/government/uploads/system/uploads/attachment_data/file/749117/Bulletin_11-2018_Hi_Res.pdf - the news story is a bit simplified, and I think it's wrong that the setting happened after takeoff, the main trigger is described in the section "Incorrect setting of the FD guidance modes prior to takeoff"

Why would a descent of 500 ft in 18 seconds be considered a "drop"?  That's a bit over 1600 FPM: a steep descent, but hardly a drop.

@quietflyer Or in the Jordan River valley, parts of which are over 1,000 feet below sea level.

Quite simply, it's because sometimes you fly below sea level. There's a couple stories out there of aircraft navigation systems acting a bit odd due to their flight below sea level. For example, there is this one, which involves a C-130 landing on an airfield that is 1,210 feet below sea level.
There are numerous areas in the world that are below sea level. For example, flying at sealevel above Laguna Salada, Mexico means you are still 30 feet above the ground! It's added as an additional option to cover a small set of scenarios. A pilot is certainly not expected to select the incorrect altitude setting, as this is something that you should always doublecheck. Beyond that, he should not have enabled his autopilot so soon after takeoff (unless 1,500 was his cruising altitude).

It's more than acceptable to engage your autopilot at 1,500 feet in most planes.  Most of the ones that I fly specify 1,000 feet as the minimum, and in busy airspace or when you want to devote more attention to avoiding weather, it can even be a good idea.

You can use the autopilot as soon as 400 ft AGL on the 737NG, and on the A320, its 100 ft AGL! (On the A320 though, you have to be already using the flight director/SRS, so you know it's going do something sensible)

Regarding the people who says "it's acceptable to engage autopilot at 1,500 feet," - what is acceptable and what is recommended, are two different things. I can tell you first hand commercial customers are typically told to climb manually,  and engage autopilot at altitude. Whether they heed that advice is to their discretion. My instructor certainly would never have allowed me to.

@Matt No way on an airliner, since a jet climb is through RVSM airspace, it would be illegal for you to hand-fly to altitude. Airline operations would expect you to be on AP by 1000 ft AGL, and there are cases like RNP AR which require you to be on AP as soon as 400 ft AGL.

Yeah, the Jordan River Valley is an interesting place. It's interesting to be on a mountain looking 1,000 feet down at the surrounding area while yourself still being below sea level.

The answer is no you would never depart with the altitude preselect at 0.  You would normally set the initial level off altitude provided in your departure clearance, or an initial level off altitude indicated in the SID, or you might set a level off altitude that you have decided to use as a level off for the single engine profile in the event of engine failure, based on the terrain around the airport.  All would be at least a 1000 ft above the airport at minimum.
You dial up that altitude in the AP's altitude preselect window, usually during the configuration setup during or after the crew departure briefing.   When you are airborne, you will normally select the autopilot's speed hold mode (on the -8 AP it's called IAS mode) which makes the flight director command a pitch attitude to hold bug speed, then when you pass 400 feet or above, you engage A/P in IAS mode and it will pitch the plane to hold the bug speed you select during the departure.  In IAS mode, whether you climb or descend depends on the power setting, so when you are climbing and descending in IAS mode you regulate the climb/descent with power while the AP pitches to hold the speed.  You can also use VS mode, or vertical speed mode, and the AP will now pitch to a preset vertical climb or descent rate.
With those two vertical modes, the incident description doesn't make a lot of sense as they would have had to have selected VS mode and dialed in a descent rate, or been in IAS mode and cut the power.  However...
The -8 400 has coupled VNAV, which adds an extra wrinkle.  With VNAV engaged, the autopilot will pitch to climb or descend to an altitude preselect independently of IAS or VS mode, instead of just hold an airspeed or vertical speed, until a preselect altitude is captured. Coupled VNAV can also follow step climbs in a SID or step descents in a STAR if that is programmed into the FMS.
So my guess here is that they forgot to set the altitude preselect for their planned departure level off altitude, then engaged the autopilot in VNAV mode during the climbout, which caused the plane to dive to the existing altitude preselect of 0 feet, upon which hilarity ensued.

For certain values of "hilarity."

Nice answer.  Thanks.

"Hilarity ensues" is a line from the old Frasier sitcom tv show when he describes his plan to play a practical joke on his boss.  One of my favourite bits.

"The answer is no you would never depart with the altitude preselect at 0." It's entirely possible for 0 MSL to be more than 1,000 feet above field level. This is possible by the Dead Sea, for example.

@JohnK: The phrase may have been popularized by Frasier, but it is far older than that. Here's an example from 1872: https://books.google.com/books?id=1nBgZNraDdkC&pg=PA317&dq=%22hilarity+ensues%22&hl=en&sa=X&ved=0ahUKEwjGp_6HmsreAhUT7Z8KHfBQDLsQ6AEINDAC#v=onepage&q=%22hilarity%20ensues%22&f=false

@reirab  well yeah I could also dig a mile deep pit and put an airport there, but we are talking about the normal world here, not outliers like that.

Besides flying below sea level, a situation for low and negative altimeter readings is use of QFE altimetry. (see also What is wrong about this interpretation of QFE and QNH?)
QFE altimetry which was used by some airlines (gone by now), and is still in use in Russia (undergoing phase-out), is where you set the altimeter's reference pressure so that it reads zero at some ground reference point, typically the runway threshold.
So if the runway is above you, on terrain, you will fly at a negative height and your altimeters will show as such.


text:
Can a change in altimeter setting make you change the airspace?
If you are at 17999 MSL VFR, you twist in 29.92 into the altimeter, and it raises your altitude to above 18000, did you bust the class A?
Then theoretically, could you be in class A airspace with a local altimeter setting if there is a very low pressure?

I'm only a helicopter pilot so anything above a couple of thousand feet to me involves sitting in comfort with someone else doing the flying ;), hence the comment rather than a definitive answer.  I believe that you cannot accidentally do this since Class A begins at the lowest flight level above the transition altitude, i.e. there is no FL18.  You switch to 29.92 when passing 18000 but, the 500 feet between this and FL185, known as the translation layer, exists to provide that buffer between QNH (or whatever setting you might be on) and class A.

91.121b has minimums for the lowest flight levels due to the altimeter setting.

No, fiddling with your altimeter setting doesn't change the airspace you are flying in. So you can't bust the class A above 18 000 ft by changing the setting. The top and bottom boundaries of the airspace are defined by applying the altimeter setting local to that airspace.
Conversely, and perhaps more important, you can't weasel out of a busted airspace by quickly changing your altimeter setting.
Your transponder sends your altitude as if your altimeter was set to 1013.25 hPA / 29.92 inHg, regardless of what your actual setting is. The ATC system will do the conversion according to their setting, so you can't fool them.
Note that the altitude resolution of your transponder is either 100 feet (Mode C) or 25 feet (Mode S with appropriate altimeter encoder), so 1 feet difference doesn't really matter.

It may we worth adding that if you are flying at the vertical border between two airspaces, the airspace class of the least restrictive airspace applies. Say, if the airspace is class C above 5000ft and class E belof 5000ft, if you are flying at 5000ft, you are legally in class E airspace

@J.Hougaard  That is not necessarily true.  There are instances where the underlying airspace stops just short of the top altitude, in your case 5000.  on sectional charts it is denoted as a "-" before the altitude.  Check out class D airspace surrounding KSDL.  That airspace extends up to but does not include 4,000 MSL.

Then theoretically, could you be in class A airspace with a local altimeter setting if there is a very low pressure?

Yes.
The answer lies in the FARs.  Class A airspace starts at 18,000' MSL.  When you set 29.92" Hg you are no longer flying a MSL altitude but a flight level.  The minimum flight level available is dependent on the current altimeter setting as specified in §91.121


text:
Did the Wright brothers patent their Wing Warping method in Germany?
I know that the Wright brothers patented their Wing Warping method for controlling the roll of a plane or glider in Great Britain, France, United States, Austria, Italy,... but did they obtain a patent for it in Germany?
This is the Austrian patent:

Orville and Wilbur Wright, “Austrian Patent No. 23174 – Flugmaschine”, Filed: 23 March, 1904, Beginning of patent term: 15 September, 1905, Issued: 26 February, 1906, 3 pages.

They have a German patent published in 1906 (see below) but I never found its content. I do not know what is written in it.

Orville Wright und Wilbur Wright in Dayton, (V. St. A.)
Patentschrift. Nr. 173378. Klasse 77h. Gruppe 7. Mitwagerechtem Kopfruder und senkrechtem Schwanzruder versehener Gleitflieger. Patentiert im Deutschen Reichevom 24. M rz 1904 ab. Ausgegeben den 16. Juli 1906. Berlin: Kaiserliches Patentamt, 1906, 4 pp., +illus.

Here's your patent: https://worldwide.espacenet.com/patent/search/family/000438192/publication/DE173378C It is well-known though that it was annuled in court


text:
Could Henri Farman have steered his plane with ailerons on January 13, 1908?
It is known that Henri Farman flew (over) 1 km in a circuit on January 13, 1908, and gained a 50000-franc prize. Orville Wright even witnessed Farman on November 18, 1907, attempting to make the full circle but each time touching the ground before completing the required course. The altitude of the machine dropped while the French pilot attempted some turns. (Farman did not use ailerons till after September 1908.)
The question is: Would the ailerons have made the Farman plane to steer more efficiently without losing altitude, which was critical because, for technical reasons, the plane could not fly higher that what you can see in the picture?

Henry Farman flying on January 13, 1908. The picture appeared in many French publications of the time.
The image below apparently shows that a plane loses altitude while taking turns using its ailerons. Does the airplane also go down if it is steered just with the tail?

This question is a good one but I think the formulation may bring more speculation than real answer. Perhaps linking this question to another speaking of boat turns and asking for altitude loss comparison for both types of turns (boat turns vs aileron turns) based on your example is better (I don't know, other opinions should be taken into account before editing your question)

I would first check to see if the car in the background was actually pulling him along in 1908, or maybe the aircraft engine was a bit better.
Banking does indeed reduce lift, but not as much as we may think:
If vertical lift at 0 degrees bank was 1000 lbs, a 20 degree bank at the same AOA and speed still produces: $$cosine 20 degrees × 1000 = 940 lbs$$ of vertical lift, easily corrected by slightly increasing the Angle of Attack of the wing.
The reason the plane previously could not make the turn can be explained by the relationship of thrust and drag.  Turning increases drag, which will slow the plane down.  These early aircraft were very draggy and underpowered, barely getting off the ground.
Deflecting the ailerons (really 2 halves of a speed brake) increases drag.  The bank additionally decreases lift.  With no reserve power to add, the plane goes down.
What an accomplishment the first circle must have been at that time!
But "skidding" the plane around the circle, with rudder only, presents the whole side of the plane as a dragging surface, there for is not a better solution.  The "coordinated turn" using  both rudder and ailerons produces the least amount of drag, and remains the recommended aircraft turning method to this day.
But even the best method of turning requires more power (thrust) than straight and level flight.  A few more horse power, and/or a lower drag design wins the prize.
Reducing weight and a better prop design would also be helpful.

An affirmation like this: "I would first check to see if the car in the background was actually pulling him along in 1908" makes me not to read the rest of your answer because it looks like you are joking.

Try 1 laugh, and please continue reading.  Yes, it is a joke, but pulling it with the car would actually work!

Yes ailerons would help turn more efficiently, although likely not enough to help with an airplane with such marginal power it can't get out of ground effect.
To the extent that it does bank, an airplane loses altitude because of the bank, and it doesn't matter how the bank was induced.  When it banks the lift vector tilted means the vertical vector is reduced and a lateral vector is introduced, the lateral vector being the turn (you're moving sideways and forward, in an arc).  The keep from descending the total lift has to be increased by pitching up, so an elevator input is required to keep the nose from dropping.  The process requires more power, so if power is that marginal, even a nice banked turn induced by proper aileron controls will cause a speed reduction as nose up input is added and it may start to settle toward the ground until you stop turning.
That being said, controlling turns by skidding, basically angling the engine thrust vector to the side to impart a lateral thrust component and around you go, more or less, is way more draggy although it can be done.  A skidding turn with wings level doesn't need as much elevator input but the drag of the skid is much worse and you will still slow down and without a power reserve will have to descend if already at a minimum speed.
Another factor is you may get some roll rate due to the skid itself to induce some bank into the turn. Depending on how strong the Farman's dihedral effect was, the skid may have induced a bit of roll as well to create a turn that was a combination of skidding and banking, (a bit like a car drifting while on a banked track turn).  This is massively inefficient and would make the problem of holding altitude with marginal power even worse.
Then, on top of that, controlling with rudder alone with only a bit of dihedral, there would be a substantial lag between the skid input and any rolling input, requiring a lot of lead-in with the rudder input. If you just try to turn by skidding, well you can make it around that way, but the roll becomes a problem because without ailerons you now have to skid the OPPOSITE way to return to level flight.
If you've ever tried to control and airplane with rudder and elevator alone, you discover this pretty quickly, unless the roll induced by yaw is very strong and immediate.  You skid, get some heading change and some roll, but to control the roll and keep from over-banking you have to skid the other way.  It's extremely difficult to control an airplane like this without very large amounts of dihedral (2 axis ultralights like the Quicksilver MX used large amounts of dihedral for this purpose, to be controllable with rudder and elevator alone; it was still an inefficient way to turn, but the thing is so draggy even flying straight it doesn't make that much difference).
In the case of the Farnham, there isn't a lot of dihedral so he would make a rudder input, skid sideways, getting some heading change, and, as there is some dihedral, there would likely be a gentle roll into the skid.  The drag alone from this slithering sideways would make the machine descend and require a pitch input.  But the delay between rudder input and a banking result would be several seconds and the thing would be a massive handful to fly, like trying to drive a car around a slalom course on a slick ice surface.
As I mentioned in my earlier answer, the Flyer was a handful to fly for other reasons, mainly neutral stability, but compared to its contemporaries it was incredibly controllable.  You would have to master the instability to train your eye/brain/muscle loop, but it was doable and once you mastered it (like a helicopter, also dynamically unstable but extremely maneuverable), you had yourself a very maneuverable machine.  The Farnham would have struggled around barely on the edge of control, a crash waiting to happen, although ailerons or wing warping would have helped a lot.

Yes, it is generally more efficient to enter a turn by using ailerons and rudder together than by using rudder alone, even if the poor performance of the aircraft requires the bank angle to be kept rather shallow in order to maintain altitude.


text:
Is drag the only aerodynamic force that acts upon an asymmetric body (a glider in particular) travelling through the air?
It is known that for a symmetric body, like a sphere falling through the air, the only aerodynamic force that acts upon it is the drag that points always in the opposite direction relative to its speed.
Now, an asymmetric body traveling through the air, I guess, will have a drag that makes an angle different from 180 degree with its speed.
As a glider is just an asymmetric body, it means that the only aerodynamic force acting upon it is the drag that is conventionally decomposed in what we call Lift and Drag.
Is it true that Lift and Drag applied to gliders or airplanes are just a convention, two forces that in reality do not exist, and there is only a True Drag that makes an angle of a certain value with the speed of the aircraft?

You are right, lift and drag are just convention. But the aerodynamic forces are usually grouped into one total force named "total aerodynamic force" ("résultante aérodynamique" in french).
As explain in how it flies:

drag is a component of the aerodynamic force, namely the projection onto the direction parallel to the relative wind.
lift is another component of the aerodynamic force, namely the projection onto the two directions perpendicular to the relative wind.

This convention applies to any aircraft.
For further reading, see the section about forces of how it flies.


text:
Is data on international air traffic volume in the US freely available?
I'm looking for air traffic flow data which has the passenger air traffic from countries outside of the US to every international airport in the US. Ideally, I would like to know how many passengers came from other countries to the international airports in the US but I can also use load factors. I'm only interested in free data. Any ideas would be appreciated. I'm willing to parse together a data from multiple datasets/sources.


text:
Is an AFSP registration required for rotorcraft rating?
I hold an FAA private license with ASEL and AMEL class ratings and with an instrument add-on rating. [I'm trying to be precise but, basically, I've an FAA multi/IFR PPL.]
I'd like to add a Rotorcraft-Helicopter class rating to my PPL. [I.e., get rated in an R44.]
I did my fixed-wing training with AFSP approval. (I'm a permanent resident.)
Do I need to obtain AFSP approval for my helicopter training?
According to the AFSP FAQ,

Category 3 Clarification - For flight training in the operation of aircraft with a maximum certificated takeoff weight of 12,500 lbs. or less, Candidates must obtain AFSP approval for the following three training events:
Single Engine Land (SEL)
-- Includes initial Airman's Certificate, including a private, recreational, or sport pilot certificate. If a private and/or commercial license is the candidate's initial FAA license, it is considered an initial airman's certificate and is not exempt.
Instrument Rating (IR)
Multi Engine Land (MEL)
Each of these three events requires a separate training request.
All other training events on aircraft with MTOW of 12,500 lbs. or less are exempt from AFSP regulations, including Commercial Pilot License (CPL), Airline Transport Pilot License (ATPL), and Certified Flight Instructor (CFI). These events are exempt only if the candidate holds an FAA stand-alone pilot certificate.

But, when asked, the AFSP said that I do need to get AFSP approval. (Citing no rule.) I've asked for clarification but does anyone know for sure?

What the AFSP said is directly contradictory to the FAQ, but if they say you need it, you need it.
I'd work on getting my citizenship asap if I was you, then you won't have to deal with this kind of stuff nearly as often.

The problem is that person writing the FAQ seems to not understand the category and class system for pilot certification. The interim final rule (the actual regulation) does not make the distinctions you have discussed from the FAQ. See https://www.flightschoolcandidates.gov/IFR_Alien_Pilot.pdf as Category 3 does not specify an FAA pilot certificate category or class, just MTOW of 12,500 pounds or less, which means a R-22, R-44, Sikorski/Schweizer 300, seaplanes, gyrocopters, gliders, lighter than air, etc. unless specifically exempt elsewhere.
According to https://www.flightschoolcandidates.gov/TSA-2004-19147-0337.pdf the measure that they use with regards to whether or not you are subject to the "threat assessment" is whether you are taking training that will "substantially enhance piloting skills" but then it further states that the requirements are for the "initial pilot certificate, including a private, recreational, or sport pilot certificate."
It would be very hard to argue that training towards a certificate in RH is not going to "substantially enhance piloting skills". However, when they came up with the FAQ it appears that they have ignored that there is any category of aircraft outside of that of airplane (and any class outside of land airplanes).
Adding on RH (or Instrument-Helicopter) to ASEL/ AMEL/ Instrument-Airplane would be your "initial pilot certificate" in the rotorcraft category, helicopter class, or your "instrument rating" for that category.
For the sake of not getting yourself or your flight school in potential serious trouble, you should probably apply under the Category 3. If I were in your position I would omit the "probably" of the last sentence, but I am not a government regulator and I am not an attorney, so there is always a degree of uncertainty with regards to my personal interpretation of this material.


text:
How does hydraulic actuator servo valve feedback work?
The book Aircraft Systems uses the following diagram for a mechanically controlled hydraulic linear actuator (for moving a control surface e.g. an aileron):

The following explanation is given about its operation:

As the pilot feeds a mechanical input to the flight control actuator, the summing link will rotate about the bottom pivot, thus applying an input to the servo valve. Hydraulic fluid will then flow into one side of the ram while exiting the opposite side resulting in movement of the ram in a direction dependent upon the direction of the pilot’s command. As the ram moves, the feedback link will rotate the summing link about the upper pivot returning the servo valve input to the null position as the commanded position is achieved.

However, I fail to understand this explanation given the diagram. If the mechanical signal is to the right, the summing link will slant to the right (i.e. /). I assume this is a command for the cylinder to move to the right. When the cylinder moves to the right, the feedback link causes the summing link to eventually return to the upright position (i.e. |). But when this position is achieved, the Servo Valve (SV) input is now more extended than it was before (the whole summing link is now upright but with a net shift to the right). How is this the null position that would cause the SV to close?
In search for an answer, I came across a better diagram which makes sense with a sliding sleeve of the SV (but this diagram is in my opinion not what the first figure illustrates):
]]
Is the first figure simply a poor explanation and the second figure is more representative of how these hydraulic linear actuators work?

The actuator in the first picture is a position follower system and only works as intended if the servo valve is closed as a consequence of the resulting piston displacement. The system in the picture works if the cylinder end is connected to the flight control surface (the left eyelet with the cross in it) and the piston end is connected to the structure frame. The key for understanding is the shift in the centre of rotation:

The servo valve is closed when the summing bar is vertical.
At first, the cylinder is stationary and the pilot deflects the stick, rotating the summing bar about the lower hinge and opening the servo valve. (Black dotted line).
While the pilot keeps the stick stationary, the cylinder moves in response of the servo valve opening. If the initial deflection was to the right, the piston moves to the right. The centre of rotation is now the top hinge (mechanical input).
When the summing bar becomes vertical again at the point where the pilot now holds the stick, the servo valve closes. (Red dotted line).

The second diagram does not show how the cylinder is attached and where the mechanical input is connceted to at the botttom. It could be a position follower system or a simple velocity output where input deflects the servo valve, oil starts to flow as long as the input is open, and the output end moves with a speed proportional to the input deflection. If the lower "First Mechanical Connection" is attached to the moving surface and the cylinder is free to move, we get a mechanical position follower again.
The mechanical feedback provided by the summing link was how the first generations of hydraulically operated systems worked. Nowadays controllers are electronic, not mechanical: a position transducer measures cylinder output, and adjusts servo valve input in response to the control laws programmed in the controller.

"When the summing bar becomes vertical again at the point where the pilot now holds the stick, the servo valve closes". Why would it close at that point? If the black line going from the centre circle on the summing link to the SV represents some mechanical link, would the SV not be more open now than it was before the pilot input? Unless the black line is just an indicator of "some" kind of mechanical connection between the SV and the summing link, which in reality may be e.g. rotary...

The black line is a mechanical connection. The whole actuator housing (marked Hydraulic Piston Actuator, with the servo valves mounted on top) moves towards the new position of the end of the link marked Mechanical Signalling. The elevator is connected to the moving housing bit, the piston is connected to the aircraft frame.

Sorry but is it not typically the piston that moves, and the housing (cylinder) that is stationary? I don't see clearly the relative movement of the housing, the cylinder and the piston

The cylinder and the piston move relative to each other. The arrangement can be made such that the cylinder is fixed and the piston extends, which can be beneficial for the oil pressure and return connections feeding the servo valve, but in that case the feedback system looks different, the one drawn is the most straightforward to explain. The aircraft I've worked on and wrote simulation models for did have a moving cylinder and a fixed piston.

"I assume this is a command for the cylinder to move to the right." If you assume this is a command for the cylinder to move to the left then the feedback is negative and the mechanism is homeostatic. It's a poor design because the summing link is pushed to awkward angles at the extremes of its motion, and it's a poor teaching example because the mechanical signalling is in the opposite direction from the powered action.
The second diagram has the feedback motion moving in the correct direction, assuming that you understand that the valve casing (gray) is free to move. However the valve is so carelessly drawn that it will not work. If the control shaft moves rightward far enough to admit pressurized oil into the power piston at P1, it completely blocks P2 and nothing happens, and there's no room for it to move leftward at all.
This sort of thing is very hard to get completely right. Hydraulics can be difficult and feedback usually is confusing; put them together and you are navigating a landscape littered with embarrassing and expensive surprises. Only the most senior engineers dare to actually build hydraulic boosters for aircraft controls and they do not sleep well at night.
So it should be no surprise to find that the diagrams in some textbooks are unclear, unhelpful, or just plain wrong.
Note: Homeostatic, said of a system, means that it has a preferred state and tends to return to that state when perturbed.
In this case the preferred state is where [SV] is centered and not admitting pressure to either side of the HPA. If you were to watch this mechanism in action, you would see that it only stops trying to move when the center pivot of the Summing Link is in a certain position. You could say that the device's preferred state is any in which this center pivot is at home, or you could say that it is this center pivot which is truly homeostatic.

That was my thinking as well. What do you define as "homeostatic"? Stable?

I've worked on design and operation of hydraulic actuators for aircraft controls and I usually sleep very well at night. Except tonight, after the heavy christmas lunch.

@Koyovis: I was thinking of https://en.wikipedia.org/wiki/Boeing_737_rudder_issues, see also http://www.b737.org.uk/rudder.htm.